The Mole

The Mole

• A mole is defined as the number of particles in exactly 12g of Carbon-12. This number = 6.02 x 1023 particles.

• 6.02 x 1023 is called Avagadro’s Number after Amadeo Avagardro who determined the volume of 1 mol of gas in 1811

• Memorize this number:– 1 mol = 6.02 x 1023 of something

The Mass of a Mole• Atomic Mass is the relative mass of single atom of an element in amu• Atomic masses on the periodic table can be used to determine molar

masses.• Molar Mass is the mass of 1 mole of something in grams

• If the atomic mass of carbon is 12.011 amu, then the molar mass of carbon is 12.011 g/mol

• IN SHORT: The molar mass of any element is exactly equal to its atomic mass – just change the units

Moles of Compounds

• In a mole of a chemical compound (molecule or formula unit), there is also one or more moles of each atom.

• For example:– In 1 mol of CCl2F2 (Freon), there is:

• 1 mol C• 2 mol Cl• 2 mol F

– If you have 2 mol of MgCl2, there is:• 2 mol Mg• 4 mol Cl• (Multiply the coefficient x the subscript!)

The Mole: In Short

• A mole is 6.02 x 1023 of anything, just like a dozen is 12 of anything.

• Atomic mass units (amu) is a way of comparing the sizes of different elements.

• atomic mass in amu = molar mass in g

Using the Mole• You should be able to:

1. Convert Moles to Particles2. Convert Particles to Moles3. Convert Moles to Mass4. Convert Mass to Moles5. Convert Mass to Particles6. Convert Particles to Mass

– Determine percent composition– Determine empirical formula– Determine molecular formula

• We are going to practice this a lot!!!!!

Preliminary: Moles to Mass of an Element

• Determine the numbers of moles of each element in 1 mole of CaF2.– 1 mole of Ca– 2 mole of F

• Determine the number of grams of each element in 1 mol of CaF2.

– 1 mol Ca x 40.1 g/mol = 40.1 grams Ca– 2 mol F x 18.9 g/mol = 37.8 grams F

Ca g1.40mol 1

g1.40

1

Ca mol 1

Preliminary: Moles to Mass of Molecule

• Determine the number of grams of each element in 1 mol of CaF2.

– 1 mol Ca x 40.1 g/mol = 40.1 grams Ca– 2 mol F x 18.9 g/mol = 37.8 grams F

• What is the molar mass of CaF2?

– 1 mol Ca + 2 mol F = 1 mole CaF2

– 40.1 g Ca + 37.8g F = 77.9 g CaF2

Preliminary: Mass of a Compound

To find the mass of a compound, simply add up the molar mass of each of its atoms. For example, calculate the molar mass of CO2.

2

2

CO g/mole 44.01 )mol 1

g 16.00( 2

mol 1

g 12.01

CO mole 1 O mol 2 C mol 1 -

g 16.00 O massmolar -

g 12.01 C massmolar -

Reminders!!

• Particles can mean: atoms, molecules, formula units, etc.

• Covalent compounds = molecules• Ionic compounds = formula units• Use the factor-label method!!!• Learn how to multiply and divide with

exponents on your calculator! (Check with me if you need help).

The Flow Chart

PartPiecesAtomsMoleculesIonsEtc.

Mass (g)Moles÷ Avo’s # x Molar Mass

÷ Molar Massx Avo’s #

#1 Moles to Particles

Determine the number of formula units in 4.3 moles of NaCl.1. Multiply the number of moles by the

appropriate conversion factor for Avogadro’s number. (Multiply by Avogadro’s #).

units formula 1059.2mol 1

1002.6

1

mol 3.4 2423

Practice: Moles to Particles

1. Determine the number of atoms in 2.50 mol Zn.

2. Determine the number of formula units (i.e., particles of an ionic compound) in 3.25 mol AgNO3.

3. Determine the number of molecules in 11.5 mol H2O.

Practice: Moles to Particles (x)

1. Determine the number of atoms in 2.50 mol Zn.

2. Determine the number of formula units (i.e., particles of an ionic compound) in 3.25 mol AgNO3.

3. Determine the number of molecules in 11.5 mol H2O.

NOTE: DON’T FORGET SIGNIFICANT DIGITS!!!

1.51 x 1024

1.96 x 1024

6.92 x 1024

#2 Particles to Moles

• How many moles are there in 5.64x1025 atoms of Fe?

• Multiply the number of atoms by the appropriate conversion factor for Avogadro’s number. (Divide by Avogadro’s #).

moles 6.93atoms106.02

mole 1

1

atoms 1064.5 23

25

Practice: Particles to Moles

How many moles contain each of the following?1. 5.75x1024 atoms of Al2. 3.75x1024 molecules of CO2

3. 3.58x1023 formula units of ZnCl2

4. 2.50x1020 atoms of Fe5. 3.4x1017 atoms of Na

Practice: Particles to Moles

How many moles contain each of the following?1. 5.75x1024 atoms of Al2. 3.75x1024 molecules of Co23. 3.58x1023 formula units of ZnCl24. 2.50x1020 atoms of Fe5. 3.4x1017 atoms of Na

9.55 mol

6.23 mol

0.594 mol

4.15x10-4 mol

5.64x10-7 mol

Practice

• WS #1 Moles to Particles

#3 Moles to Mass

If you have 0.0450 mol of Cr, how many grams do you have?

1. Determine the molar mass of Cr from the periodic table: 52.00 g/mol (same number as the atomic mass)

2. Multiply the number of moles by the conversion factor. (Multiply by the molar mass).

Cr g 34.2rmol 1

C g 52.0

1

Cr mol 0.0450

Practice: Moles to Mass (x)

Convert from moles to mass:1. 3.57 mol Al2. 42.6 mol Si3. 3.45 mol Co4. 2.45 mol Zn5. 8.2 mol Pb

Practice

Convert from moles to mass:1. 3.57 mol Al2. 42.6 mol Si3. 3.45 mol Co4. 2.45 mol Zn5. 8.2 mol Pb

96.3 g

1.2 x 103 g

203 g

1.60 x 102 g

1.7 x 103 g

#4 Mass to Moles

How many moles of Ca are in 525g of Ca?1. Determine the molar mass of Ca from the

periodic table: 40.08 g/mol2. Multiply the given mass (525g) by the

conversion factor. (Divide by the molar mass).

Ca mol 1.13 g 08.40

mol 1

1

Ca g 525

Practice: Mass to Moles

Convert from mass to moles:1. 25.5 g Ag2. 300.0 g S3. 125 g Zn4. 99g O5. 1.0 kg Fe

Practice: Mass to Moles

Convert from mass to moles:1. 25.5 g Ag2. 300.0 g S3. 125 g Zn4. 99g O5. 1.0 kg Fe

0.236 mol

9.355 mol

1.91 mol

6.2 mol

17.9 mol

Practice

• WS #2 Mole to Mass• WS #3 Mixed One-Step Mole WS

Multi-Step ProblemsHow many molecules are there in 24 g of FeF3?

1. Calculate the molar mass of FeF3

2. Divide the mass you have by the molar mass. This gives you the number of moles.

3. Multiply the number of moles by Avogadro's number.

How many grams are there in 7.5 x 1023 molecules of AgNO3?

1. Calculate the molar mass of AgNO3.

2. Divide the number of molecules by Avogadro’s number. This gives you the number of moles.

3. Multiply the number of moles by the molar mass.

atoms10 x 26.110 x 6.02 x 0.21 3.

mol 0.21 g/mole 112.8 g 24 2.

g 112.8 g 57.0 g 55.8 19.0) x (3 3F g) (55.8 Fe .1

2323

g 209 g/mol 169.9 x mol 1.2 3.

mol 1.2 atoms/mol 10 x 6.02 atoms 10 x 7.4 2.

g/mol 169.9 16.0) x (3 3O g) (14.0 N g) (107.9 Ag .12323

#5 Mass to Particles

Calculate the number of atoms in a 25.0 g nugget of pure gold (Au).

1. Convert mass to moles (mass ÷ molar mass)2. Convert moles to atoms (moles x 6.02 x 1023)

Au atoms 1065.7atoms/mol 10 .026 Au mol 0.127 2.

Au mol 127.0Au g/mol 196.7

Au 25g 1.

2223

Practice: Mass to Particles

• How many atoms are there in 14 g of C?

#6 Particles to Mass

What is the mass of 5.50 x 1022 He atoms?1. Convert atoms to moles (atoms ÷ 6.02 x 1023)2. Convert moles to mass (moles x molar mass)

He g 0.366 He g/mol 00.4 He mol 0.0914 2.

He mol 0914.0atoms/mol 1002.6

He atoms 1050.5 .1

23

22

Practice: Particles to Mass

• What is the mass of 7.6 x 1024 atoms of Ca?

Practice: Mass to Atoms and Atoms to Mass

Convert to Atoms1. 55.2 g Li2. 0.230 g Pb3. 11.5 g Hg4. 45.6 g Si5. 0.120 kg Ti

Convert to Mass1. 6.02 x 1024 atoms Bi2. 1.00 x 1024 atoms Mn3. 3.40 x 1022 atoms He4. 1.50 x 1015 atoms N5. 1.50 x 1015 atoms U

Practice: Mass to Atoms and Atoms to Mass

Convert to Atoms1. 55.2 g Li2. 0.230 g Pb3. 11.5 g Hg4. 45.6 g Si5. 0.120 kg Ti

Convert to Mass1. 6.02 x 1024 atoms Bi2. 1.00 x 1024 atoms Mn3. 3.40 x 1022 atoms He4. 1.50 x 1015 atoms N5. 1.50 x 1015 atoms U

3.45 x 1022atoms

6.68 x 1020 atoms

1.51 x 1024 atoms

9.77 x 1023atoms

91.3 g

2.09 x 103 g

3.49 x 10-8 g

0.226 g

5.93 x 10-7 g

4.79 x 1024 atoms

Practice

• WS #4 Two-Step Mole Conversions• WS #5 Mixed Problems w/ Naming • WS #6 Mixed Problems

Percent Composition by Mass

• The percent composition by mass tells you what percent an element is out of a compound.

• For example:– Hydrogen (H) is 11.2% of H2O by mass

H % 11.2 100% OH g 18.02

H g 2.02

massby percent 100% compound of mass

element of mass

2

The Steps: Percent by Mass

1. Find the molar mass of the elements in the compound.

2. Find the molar mass of the compound.3. Divided the molar mass of each element by

the molar mass of the compound and multiply by 100%.– (Make sure you multiply the mass of the element

by the number of moles present in the compound)

Example:

Find the percent composition of each of the elements in NaHCO3

O %15.57%10084.01

48.00

NaHCO g/mol 84.01

O g/mol 16.00 O mol 3

C %30.14%10084.01

12.01

NaHCO g/mol 84.01

C g/mol 12.01 C mole 1

H %200.1%10084.01

1.008

NaHCO g/mol 84.01

H g/mol 1.008 H mole 1

Na %37.27%10084.01

22.9

NaHCO g/mol 84.01

Na g/mol 22.9 Na mole 1

g/mol 01.84NaHCO of massmolar

3

3

3

3

3

Practice

1. Determine the percent by mass of each element in calcium chloride (CaCl2).

2. Calculate the percent by mass of each element in sodium sulfate (Na2SO4)

3. Calculate the percent composition of phosphoric acid (H3PO4).

Practice

1. Determine the percent by mass of each element in calcium chloride (CaCl2).– Ca = 36.11%, Cl = 63.89%

2. Calculate the percent by mass of each element in sodium sulfate (Na2SO4)– Na = 32.37%, S = 22.58%, O = 45.05%

3. Calculate the percent composition of phosphoric acid (H3PO4).– H= 3.08%, P = 31.61%, O = 65.31%

Practice: Percent Composition

• WS #7 Percent Composition by Mass• WS #8 Percent Composition Hard

Molecular Formula vs. Empirical Formula

• A molecular formula indicates the actual number of elements in a covalent compound.

• We call the mass (in amu) of the molecular formula the molecular mass.– Remember: we can convert molecular mass to molar mass by simply

changing the units (amu g/mol).• The empirical formula for a molecule is the smallest whole-

number ratio of the elements. It may or may not be the same as the molecular formula.

• We call the mass (in amu) of the empirical formula the formula mass.– Remember: we can convert the formula mass to a molar mass simply by

changing the units (amu g/mol).

A Few Key Conceptual Points:

• The molecular formula can be the same as or different from the empirical formula.

• Ionic compounds ONLY have empirical formulas (lowest whole number ratios) because of their crystal lattice structure.

• Only covalent compounds can have molecular formulas that differ from their empirical formulas.

• For covalent compounds, it is the molecular formula that occurs in real life.

Example: Empirical vs. Molecular

• The molecular formula for glucose (a product of photosynthesis) is C6H12O6.

• The empirical formula is CH2O.

• The molecular formula for glucose is 6 times the empirical formula.

2 New Skills to Learn

1. How to calculate empirical formula from percent composition.

2. How to calculate molecular formula from empirical formula and molar mass.

#1 Calculating Empirical Formula from Percent Composition

1. Assume you have a 100 g sample and convert your percentages to masses.

2. Divide the masses that you have from step #1 by their molar masses to determine the mole ratios.

3. Calculate the simplest ratio by dividing each mole amount from #2 by the smallest mole amount that you calculated from #2.

4. Convert to the smallest whole-number ratio by multiplying all molar amounts by a common multiple. (No parts of atoms rule).

Example: Calculating Empirical Formula from Percent Composition

• Example: Imagine that you know the percent composition of a compound, ethane.

• Problem: Determine the empirical formula from the percent composition:

• C = 79.9%• H = 20.1%

The Steps

1. Assume you have 100 g– 79.9% C = 79.9g C and 20.1% H = 20.1g H

2. Divide the mass of each element by its molar mass.– (79.9g ÷ 12.01 g/mol) = 6.65 mol– (20.1g ÷ 1.01g/mol) = 19.9 mol

3. Divide the molar amounts from step 2 by the smallest molar amount (6.65 mol).– (6.65 mol C ÷ 6.65 mol) = 1 C– (19.9 mol H ÷ 6.65 mol) = 3 H

4. Determine the ratio of elements and convert to whole numbers by multiplying by a common multiple if necessary.– Therefore, the empirical formula for ethane is CH3

Practice: Empirical Formula from Percent Composition

• Determine the empirical formula for the following compounds based on the percent composition given.

1. 36.84% nitrogen, 63.16% oxygen2. 35.98% aluminum, 64.02% sulfur3. 81.82% carbon, 18.18% hydrogen

N2O3

Al2S3

C3H8

Skill 2: Molecular Formula from Empirical Formula

• A covalent compounds molecular formula is some integer multiple of the empirical formula.– i.e., Molecular formula = n(empirical formula)• Where n= natural numbers (1, 2, 3, etc.)

• The molecular formula is found by comparing the ratio of the molecular mass to the formula mass.

Example

• Molecular mass of acetylene = 26.04 amu• Formula mass of acetylene = 13.02 amu• The ratio of molecular mass to formula mass:– (26.04 amu)/(13.02 amu) = 2

• Therefore: the molecular mass is 2x the formula mass, so the molecular formula has 2x the number of each elements found in the empirical formula.

The Steps

Note: You will be given the molecular mass!! Otherwise, you can’t figure this out!!1. Determine the empirical formula of the compound.2. Determine the formula mass of the empirical

formula.3. Divide the molecular mass by the formula mass.– (molar mass) ÷ (formula mass) = ????

4. Multiply the subscripts of each element in the empirical formula by the number determined in step 3. You’re done.

Example:

• Determine the molecular formula for succinic acid, which is 40.68% C, 5.08% H, and 54.24% O. It has an experimentally determined molecular mass of 118.1 amu.

1. Find the empirical formula.– C2H3O2

2. Calculate the formula mass of the empirical formula.– 59.05 amu

3. Divide the molecular mass by the formula mass.– (118.1 amu) ÷ (59.05 amu) = 2

4. Multiply the empirical formula by the whole number you found in step 3.– 2(C2H3O2) = C4H6O4

Practice Problems

1. A compound with a molar mass of 60.01 g/mol is 46.68% N and 53.32% O. What is its molecular formula?

2. The experimentally determined molar mass of an unknown compound is 110.0 g/mol, and it is 65.45% C, 5.45% H, and 29.09% O. What is its molecular formula?

C6H6O2

N2O2

Practice

• WS #9 Empirical & Molecular Formula• WS #10 Empirical & Molecular

Hydrates• Some salts have the ability to bind water molecules

within their lattice structure. These compounds are known as hydrated salts or hydrates, written as CoCl2 • 6H2O. The salt with no water in the crystal lattice is called an anhydrous salt CoCl2.

• Heating a hydrated salt will drive off the water molecules drying out the crystal and forming the anhydrous salt.

• Anhydrous salts will absorb water from the air. They are used to keep products dry by inserting little packets into shoe boxes, cameras etc.

Formula For a Hydrate

• In the formula for a hydrate, the number of water molecules associated with each formula unit of the compound is written following a dot.

• Example:– Sodium carbonate decahydrate – Na2CO3 10H∙ 2O

• The number of water molecules associated with the formula unit in the compound are indicated in the prefix.

• The mass of water associated must always be included in the molar mass

% Composition of Hydrates% Composition of hydrate tells you how much of the hydrate is salt, and how much

is water.

It’s just like the other percent composition calculations.

1. Calculate the molar masses for the anhydrous salt, the water and the total mass.

2. Divide the molar mass of each part (salt, water) by the total molar mass and multiply by 100%. Done.

• % = Part / whole X 100

– % H2O = (mass H2O) ÷ (total mass) x 100

– % salt = (mass salt) ÷ (total mass) x 100

Practice Problem

1. What is the % composition of CaSO4 · 2H2O?

H20 = 21%

CaSO4 = 79%

Analyzing a Hydrate

• To determine the formula for a hydrate you must determine the moles of water associated with each mole of hydrate.

• The water is boiled off, and the before and after masses of the compound are compared. The difference is the mass of the water in the compound.

• The differences in grams are converted to molar masses and the ratio of the formula unit to water is determined.

Hydrate Example• You start with 5.00 g of BaCl2 · xH2O. After heating there is only 4.26 g of BaCl2.

What is the formula for the hydrate?

1. Find the difference between the two masses. This is the mass of water.– 5.00g – 4.26g = 0.74g H2O

2. Divide the mass of each (salt, water) by their molar masses. This gives you the # of moles.– BaCl2 : (4.26g)/(208.23g/mol) = 0.0205 mol BaCl2– H2O: (0.74g)/(18.02g/mol) = 0.041 mol H2O

3. Divide the moles of water by the moles of salt. This gives you the mole ratio of the hydrate. You’re done. – (0.041 mol)/(0.0205 mol) = 2/1 (Note: if you get decimals, round).– So there are 2 moles of water per mole of barium chloride– Barium chloride dihydrate = BaCl2 2H∙ 2O

Practice Problems

1. During a lab, 1.62g of CoCl2 · xH2O is heated. After heating, 0.88g CoCl2 remains. What is the formula for the original hydrate?

CoCl2 · 6H2O

Practice

• WS #11 Hydrate Problems• Mole PRACTICE Quiz• Mole Quiz• Mole Test Review